We go by Option method for this type of problem

Let number of rows = 3

Let the number of students in the 3 rows be n - 3, n and n + a

⸫  (n - 3) + n + (n + 3) = 630

=  3n=630 or n=210

⸫  It is possible to seat the children in 3 rows.

Let number of rows = 4

Let the number of students in the 4 rows be n, n + 3, n + 6 and n + 9.

Then, n + (n + 3) + (n + 6) + (n + 9) = 630

=  4n + 18 = 630  ]  4n = 630 - 18

=  4n = 612 or n = 153

⸫  Such an arrangement is possible.

Let number of rows = 5

Let the number of students in the 5 rows be n - 6, n – 3n, n + 3 and n + 6

Then, (n - 6) + (n – 3) + n + (n + 3) + (n + 6) = 630

=  5n = 630 or n = 126

⸫  Seating the children in 5 rows is also possible.

Let number of rows = 6

Let the number of students in the 6 rows be n, n + 3, n + 6, n + 9, n + 12 and n + 15.

Then.  n + (n + 3) + (n + 6) + (n + 9) + (n + 12) + (n + 15) = 630

=  6n + 45 = 630  ]  6n = 630 - 45

=  6n = 585 or n = 97.5

Which doesn't turn out to be an integer.

Thus, arrangement in 6 rows is not possible.

Hence, option (d) is correct.